Q133 / 160 · TS EAMCET 2019 (04 May Shift 1) · Physics
If a planet of mass $6.4 \times 10^{23} \mathrm{~kg}$ can be compressed into a sphere such that the escape velocity from its surface is $8 \times 10^4 \mathrm{~m} / \mathrm{s}$, then what showld be the radius of the sphere? (Gravitational constant, $G=6.6 \times 10^{11} \mathrm{Nm}^2 \mathrm{~kg}^2$ )
- A$40.4 \mathrm{~km}$
- Correct answerB$13.2 \mathrm{~km}$
- C$20.4 \mathrm{~km}$
- D$6.8 \mathrm{~km}$
Solution
As, we know, the escape velocity of a planet is given as
or
$$
\begin{aligned}
& v_e=\sqrt{\frac{2 G M}{R}} \\
& v_e^2=\frac{2 G M}{R}
\end{aligned}
$$
Radius of sphere $=\frac{2 G M}{v_e^2}$
where, $M=6.4 \times 10^{23} \mathrm{Kg}, v_e=8 \times 10^4 \mathrm{~m} / \mathrm{s}$
and $G=6.6 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 \mathrm{~kg}^{-2}$
So, $R=\frac{2 \times 6.6 \times 10^{-11} \times 6.4 \times 10^{23}}{64 \times 10^8}=13.2 \times 10^3$
$\Rightarrow R=13.2 \mathrm{~km}$
Hence, option (2) is correct.
Hence, the correct answer is option: 'B'.